Bases, dimension and linear transformations

One vector can have different coordinate lists

The vector v=(5,1) can be built as 5e₁+e₂ using e₁=(1,0), e₂=(0,1), or as 2b₁+b₂ using b₁=(2,1), b₂=(1,−1). Relative to the horizontal and vertical unit vectors, its coordinates are (5,1). Relative to the ordered pair B=(b₁,b₂), its coordinates are (2,1). The vector has stayed in the same place; the two lists describe different building instructions.

The pair B reaches every target in R², and no target has two different recipes using that pair. These are exactly the spanning and independence properties established in Vector spaces 123. A list with both properties is called a basis. Its coordinates are useful because there is one recipe for each vector.

Now let T move each vector by the rule T(x,y)=(x+2y,y). The same v is sent to (7,1). This time the vector really has changed. We will keep a change of coordinates separate from an action on a vector, then connect both calculations using matrices.

Preparation: finite spans and independence from Vector spaces 123, and matrix-vector multiplication and reversible elimination from Matrices 122. All spaces and scalars here are real. The main arguments concern finite-dimensional spaces. For a first reading, follow the plane examples and coordinate diagrams; return to the basis-length and kernel-dimension proofs after the concrete computations. The polynomial section explains its own coefficient rule.

A basis spans the stated space without redundancy

A finite ordered basis B=(b₁,…,bₙ) of a vector space V is a list of vectors in V that spans V and is linearly independent. Every v∈V can therefore be written in exactly one way as v=c₁b₁+⋯+cₙbₙ. Spanning gives existence; subtracting two recipes and using independence gives uniqueness.

Both conditions refer to a stated space. The single vector (1,0) is independent in R² but does not span R². The list (1,0),(0,1),(1,1) spans R² but has the relation (1,0)+(0,1)−(1,1)=0. Neither list is a basis of R². A basis cannot contain zero or a repeated member.

The standard ordered basis of Rⁿ consists of e₁,…,eₙ, where eⱼ has entry 1 in position j and 0 elsewhere. The coefficients in v=Σvⱼeⱼ are its ordinary coordinate entries. A different ordered basis need not consist of perpendicular or unit-length vectors. Independence and spanning are the requirements.

The zero space {0} has the empty basis: the empty sum gives its only vector, and the empty list is independent. A one-member list containing 0 is dependent and is not that basis. Hereafter, the displayed coordinate matrices use positive dimensions; this convention about the zero space still matters in the general definitions.

Why every finite basis has the same length

A space is finite-dimensional when some finite list spans it. Starting with that list, discard a member whenever it is a combination of the others. Each removal keeps the same span. The process stops after finitely many removals with an independent spanning list, so a basis exists. Starting with an independent list instead, add members of a finite spanning list only when they are outside the span so far. This extends the independent list to a basis.

The key comparison is that an independent list cannot be longer than a finite spanning list in the same space. To see why, let u₁,…,uᵣ be independent and w₁,…,wₛ span V. Express u₁ using the w's. Some coefficient is nonzero, so solve for that w and replace it by u₁; the new list still spans V. At step k, express uₖ using the earlier u's and the remaining w's. A remaining w must have a nonzero coefficient, or uₖ would be a combination of the earlier u's, contradicting independence. Replace that w. There are only s original slots to replace, so r≤s.

Apply this comparison to two bases B and C. Since B is independent and C spans, the length of B is at most the length of C. Reversing their roles gives the reverse inequality. Their lengths are equal. We can therefore define dim V, the dimension of V, as the number of vectors in any basis. In particular, dim{0}=0 and dim Rⁿ=n.

If dim V=n, any independent list of n vectors in V is a basis, and any spanning list of n vectors is a basis. An independent shorter list can be extended; a dependent spanning list can be shortened. The condition that the vectors belong to V is essential. Counting n vectors without that condition, or without either independence or spanning, proves nothing by itself.

Dimension belongs to the space, not its ambient notation

Consider W={(X,Y,Z)∈R³:X+2Y−Z=0}. Every member has the form (−2s+t,s,t)=s(−2,1,0)+t(1,0,1). These two generators are independent: their second and third coordinates force s=t=0 in a zero combination. They form a basis of W, so dim W=2 even though each vector has three ambient coordinate entries.

The space P₂ consists of all real polynomials a+bt+ct², including the zero polynomial. The ordered list (1,t,t²) spans P₂, and equality of polynomial coefficients proves its independence. Thus dim P₂=3. Degree at most 2 counts three coefficient positions; a polynomial need not have degree exactly 2 to belong to P₂.

For a finite list of coordinate vectors, place them in columns and use elimination to locate the pivot columns. Those positions select a basis of their span from the original columns. The reduced columns generally span a different subset of the original ambient coordinate space. To find a basis of a homogeneous solution space, use its free-parameter generators instead, as in the description of W.

A subspace U of a finite-dimensional V also has a finite basis. Start with the empty list and add a vector of U outside its current span whenever that span is not yet U. Each addition keeps the list independent. The length bound from the preceding section prevents this list from becoming longer than a basis of V, so the process must stop with a finite independent spanning list for U. Thus dim U is defined and dim U≤dim V. If the dimensions are equal, a basis of U is an independent list of dim V members in V and hence spans V; therefore U=V. For example, the plane W is a proper subspace of R³ because its dimension is 2 rather than 3.

Coordinates use a named and ordered basis

For B=(b₁,…,bₙ), write [v]ᴮ=(c₁,…,cₙ)ᵀ when v=Σcⱼbⱼ. The superscript B names the basis; the symbol ᵀ says that the listed numbers form a column. Coordinates are real numbers attached to ordered positions. Reversing the order of B reverses the corresponding positions, even though the set of basis vectors has not changed.

When the basis vectors are given in ambient coordinates, let Pᴮ be the matrix with those vectors as columns. Reconstructing a vector means multiplying: v=Pᴮ[v]ᴮ. Finding its coordinates means solving Pᴮc=v. Since B is a basis of V, the solution is unique for every v∈V. A vector outside V has no such coordinate list. The matrix can be rectangular when V is a subspace of a larger coordinate space.

For B=((2,1),(1,−1)), Pᴮ=[[2,1],[1,−1]]. Solving 2c₁+c₂=5 and c₁−c₂=1 gives [v]ᴮ=(2,1)ᵀ for v=(5,1). With C=((1,1),(0,1)), the equations d₁=5 and d₁+d₂=1 give [v]ᶜ=(5,−4)ᵀ. Reconstructing gives 2(2,1)+(1,−1)=(5,1)=5(1,1)−4(0,1).

The coordinate assignment preserves sums and scalar multiples. If v has coefficient list c and w has coefficient list d, then v+w has list c+d; uniqueness identifies it as [v+w]ᴮ. Similarly, [av]ᴮ=a[v]ᴮ. Thus using basis coordinates gives a linear correspondence between V and Rⁿ, including a two-dimensional plane such as W and its two coordinate numbers.

A linear map preserves combinations

A linear map T:V→W between real vector spaces satisfies T(u+v)=T(u)+T(v) and T(av)=aT(v) for every u,v∈V and every a∈R. Equivalently, T(au+bv)=aT(u)+bT(v) for all those choices. The domain V and codomain W are part of the map's specification; every output must belong to W.

For T:R²→R² given by T(x,y)=(x+2y,y), expanding the coordinate expressions verifies both laws. It sends (5,1) to (7,1), and it sends the two basis vectors (2,1),(1,−1) to (4,1),(−1,−1). Therefore T(2b₁+b₂)=2T(b₁)+T(b₂). The picture shows the same weighted construction before and after the map.

Every linear map sends 0 to 0, since T(0v)=0T(v)=0. The translation F(x,y)=(x+1,y) fails this test: F(0,0)=(1,0). It is an affine map with a nonzero offset. The rule G(x,y)=(x²,y) does send zero to zero, but G(2,0)=(4,0) differs from 2G(1,0)=(2,0). Passing the zero test is necessary and is not sufficient.

A graph being a straight line is not the definition used here. A nonzero constant term breaks linearity under the usual vector-space operations. Conversely, a linear map can collapse a direction or map between spaces of different dimensions. Preservation of combinations is the condition to check.

Images of a basis determine the whole map

Suppose B=(b₁,…,bₙ) is a basis of V. Choose any vectors w₁,…,wₙ in W and prescribe T(bⱼ)=wⱼ. For a vector v with its unique expansion v=Σcⱼbⱼ, define T(v)=Σcⱼwⱼ. The definition is unambiguous because v has only one coefficient list. Adding or scaling coordinate lists shows that the resulting T is linear.

This is the only linear map with those basis images: any linear map must send Σcⱼbⱼ to ΣcⱼT(bⱼ). The output vectors wⱼ need not be independent or span W. If two basis directions acquire the same image, or one acquires zero, the linear map still exists; it may lose information.

A redundant generating list needs an extra consistency check. If g₃=g₁+g₂, a linear map must satisfy T(g₃)=T(g₁)+T(g₂). Giving those three generators arbitrary, incompatible images would assign two different outputs to the same input. Independence is what permits completely free choices of basis images.

In standard coordinates of Rⁿ, this explains the columns of a matrix transformation. Since x=Σxⱼeⱼ, its image is T(x)=ΣxⱼT(eⱼ). Placing T(eⱼ) in column j makes matrix multiplication perform exactly that sum. The same idea works for nonstandard input and output bases once their coordinate conventions are stated.

Each matrix column records one basis image

Let B=(b₁,…,bₙ) be a basis of V and C=(c₁,…,cₘ) a basis of W. The matrix [T]ᶜ←ᴮ has column j equal to [T(bⱼ)]ᶜ. Its n columns correspond to input basis vectors; its m rows correspond to output basis coordinates. The arrow in the notation reads 'C coordinates from B coordinates'.

If [v]ᴮ=(a₁,…,aₙ)ᵀ, linearity gives T(v)=ΣaⱼT(bⱼ). Taking C coordinates gives [T(v)]ᶜ=[T]ᶜ←ᴮ[v]ᴮ. Thus the matrix acts on a coordinate column. Its output is another coordinate column, which must be reconstructed using C to recover the output vector itself.

Use T(x,y)=(x+2y,y), B=((2,1),(1,−1)) and C=((1,1),(0,1)). We have T(b₁)=(4,1)=4c₁−3c₂, and T(b₂)=(−1,−1)=−c₁+0c₂. Hence [T]ᶜ←ᴮ=[[4,−1],[−3,0]]. For v=(5,1), multiply it by [v]ᴮ=(2,1)ᵀ to obtain (7,−6)ᵀ. Reconstructing 7c₁−6c₂ gives (7,1), the same output as the original formula.

The standard matrix [[1,2],[0,1]] and the matrix [[4,−1],[−3,0]] describe the same T with different coordinate conventions. They accept and return different coordinate lists. Mixing a B-coordinate input with the standard matrix usually gives the wrong interpretation, even when both arrays have two entries.

Keep the kernel, image and codomain distinct

The kernel is ker T={v∈V:T(v)=0}, a subspace of the domain. The image is im T={T(v):v∈V}, a subspace of the codomain. Linearity proves closure for both: zero maps to zero, and combinations of inputs produce the corresponding combinations of outputs. The image need not equal the whole codomain.

A map is injective, or one-to-one, when equal outputs force equal inputs. For a linear map this is equivalent to ker T={0}: T(u)=T(v) is the same as T(u−v)=0. It is surjective, or onto W, when im T=W. Projection P:R²→R², P(x,y)=(x,0), has kernel the y-axis and image the x-axis. It is neither injective nor onto R².

Now consider E:R²→R³ given by E(x,y)=(−2x+y,x,y). The second and third output coordinates recover the input, so E is injective. Its image is exactly W={(X,Y,Z):X+2Y−Z=0}: every output satisfies the equation, and every member (−2s+t,s,t) is E(s,t). E is not onto R³ because, for example, (1,0,0) is not in its image.

Specify instead a map with the same input space and formula but codomain W. It is onto that codomain and is still injective. The ambient three-entry output has not changed; the question 'onto what space?' has. A matrix in coordinates of a basis of W has two rows, while a standard-coordinate matrix into R³ has three rows. The two row counts describe different codomain specifications.

Count the lost and retained directions

For a linear map with a finite-dimensional domain, dim V=dim ker T+dim im T. The image dimension is called the rank of T, and the kernel dimension is its nullity. The identity counts independent input directions that vanish and independent output directions that remain. It is not a count of the number of vectors in either space.

To justify it, take a basis k₁,…,kᵣ of ker T and extend it to a basis (k₁,…,kᵣ,u₁,…,uₛ) of V. Applying T to a basis expansion removes the k terms, so T(u₁),…,T(uₛ) span im T. If a combination ΣaⱼT(uⱼ) is zero, then Σaⱼuⱼ belongs to ker T. Expressing it using the k's gives a relation in the extended basis, forcing every aⱼ to zero. The s images are independent and form a basis of im T. Thus dim V=r+s.

Projection of R² onto its x-axis has nullity 1 and rank 1, giving 2=1+1. The embedding E into R³ has nullity 0 and rank 2, giving 2=0+2. The same formula with codomain W also has rank 2; changing the codomain from R³ to W changes surjectivity, not the image or its dimension.

In a coordinate matrix [T]ᶜ←ᴮ, solving the homogeneous equation finds kernel coordinates in B; reconstruct them using B to obtain actual kernel vectors. Original pivot columns give a basis of the matrix column space in C coordinates; reconstruct those columns using C to obtain an image basis in W. The pivot count is dim im T because the coordinate correspondence preserves independence and spans. The next lesson develops rank and invertibility further.

A polynomial map has basis columns too

Define D:P₂→P₁ by D(a+bt+ct²)=b+2ct. This is the usual derivative rule for these polynomials; it can also be checked directly as a rule on coefficients. Adding two coefficient triples or multiplying one by a real scalar commutes with this rule, so D is linear. No limiting calculation is needed for this example.

Use the ordered input basis (1,t,t²) and output basis (1,t). Their images are D(1)=0, D(t)=1 and D(t²)=2t, so the coordinate columns are (0,0)ᵀ, (1,0)ᵀ and (0,2)ᵀ. The resulting matrix is [[0,1,0],[0,0,2]]. Acting on (2,3,1)ᵀ gives (3,2)ᵀ, which represents 3+2t, the image of 2+3t+t².

D(p)=0 forces b=c=0, so ker D is the space of constant polynomials, with basis (1). Every polynomial u+vt in P₁ is reached by p=ut+(v/2)t². Thus im D=P₁ and D is onto P₁, but it is not injective. The dimensions are 3=1+2. Polynomials differing by a constant have the same image.

If the derivative formula is specified instead as a map P₂→P₂, its image is still the subspace P₁ inside P₂. It is no longer onto its codomain: t² has no preimage in P₂. Its standard coordinate matrix has a third, zero output row. Changing the polynomial bases changes the coordinate matrix without changing which polynomials the rule sends to which outputs.

Coordinate changes and compositions have directions

A coordinate-change matrix Sᶜ←ᴮ converts [v]ᴮ into [v]ᶜ for the same vector v. Its columns are [bⱼ]ᶜ, so it is the matrix of the identity map with input basis B and output basis C. It need not be the identity array when B and C differ. For our plane bases, b₁=2c₁−c₂ and b₂=c₁−2c₂, giving Sᶜ←ᴮ=[[2,1],[−1,−2]]. It sends (2,1)ᵀ to (5,−4)ᵀ, the two coordinate lists of v=(5,1).

To compute a coordinate change, solve for the destination-basis coordinates of each source-basis vector. The reverse conversion undoes that operation. For bases of the whole Rⁿ, one can also write Sᶜ←ᴮ=(Pᶜ)⁻¹Pᴮ: first reconstruct with B, then solve for C coordinates. This square-matrix inverse notation is optional here. For a plane basis in R³, its three-by-two ambient column matrix has no square inverse; solve the consistent coordinate system instead.

For linear maps T:U→V and R:V→W, the composition R∘T means apply T first, then R. With bases B of U, C of V and F of W, its matrix is [R∘T]ᶠ←ᴮ=[R]ᶠ←ᶜ[T]ᶜ←ᴮ. The output coordinates of the first factor applied must match the input coordinates of the next. A matching array size alone does not identify a matching intermediate basis.

For the standard-coordinate shear T(x,y)=(x+2y,y) and swap R(x,y)=(y,x), applying T then R sends (1,2) to (5,2), then (2,5). Reversing the order sends (1,2) to (2,1), then (4,1). The matrices are [[0,1],[1,2]] for R∘T and [[2,1],[1,0]] for T∘R. The order matters, as it did for matrix multiplication in Matrices 122.

Predict, then change a basis

Choose a map and a target input, then select its ordered input and output bases. Predict which quantities will change. For the same map, target and codomain, the coordinate lists and matrix can change, while the input vector, output vector, kernel, image, rank and nullity stay the same. The reconstruction checks display both coordinate routes.

Compare the two plane-embedding cases or the two derivative cases to change the declared codomain while keeping the formula. Inspect the stated space, not just the ambient row count, before predicting whether the map is onto. The controls use exact finite examples; they illustrate the all-real statements proved above. Determinants, eigenvectors and numerical conditioning come later.

Worked examples

Try it yourself

1. Is ((1,0),(0,1),(1,1)) a basis of R²?

   Show the answer and reasoning

   No.

   It spans R² but has the nontrivial relation (1,0)+(0,1)−(1,1)=0, so it is dependent.

2. Can two independent vectors in R³ form a basis of R³? Can they form a basis of their span?

   Show the answer and reasoning

   They cannot form a basis of R³; they do form a basis of their own span.

   A basis of R³ has three members. Their own span is generated by them, and independence supplies the other basis condition, giving dimension 2.

3. What is a basis of {0}, and what is its dimension?

   Show the answer and reasoning

   The empty list is a basis, and the dimension is 0.

   The empty sum gives its only vector. A list containing one zero vector is dependent and is not a basis.

4. Why is dim P₂ equal to 3 rather than 2?

   Show the answer and reasoning

   The basis (1,t,t²) has three members.

   Every polynomial of degree at most 2 has three coefficient positions, and equality of coefficients proves that the listed powers are independent.

5. If [v]ᴮ=(2,1)ᵀ for B=(b₁,b₂), what are the coordinates in the reversed basis (b₂,b₁)?

   Show the answer and reasoning

   (1,2)ᵀ.

   The vector v=2b₁+b₂ is also 1b₂+2b₁. The coefficient positions follow the basis order.

6. Does (1,0,0) have coordinates in the basis ((−2,1,0),(1,0,1)) of W?

   Show the answer and reasoning

   No.

   The last two coordinates would force both coefficients to zero, producing first coordinate 0 rather than 1. The vector lies outside W.

7. G(x,y)=(x²,y) sends zero to zero. Is it linear?

   Show the answer and reasoning

   No.

   G(2,0)=(4,0), whereas 2G(1,0)=(2,0), so scalar multiplication is not preserved.

8. Suppose g₃=g₁+g₂. Can a linear map have T(g₁)=T(g₂)=0 but T(g₃)≠0?

   Show the answer and reasoning

   No.

   Linearity forces T(g₃)=T(g₁)+T(g₂)=0. Arbitrary image prescriptions are valid for a basis, not for a redundant list without consistency checks.

9. With T(x,y)=(x+2y,y), B=((2,1),(1,−1)) and C=((1,1),(0,1)), what is column 2 of [T]ᶜ←ᴮ?

   Show the answer and reasoning

   (−1,0)ᵀ.

   The second input basis vector is (1,−1). Its image is (−1,−1)=−c₁+0c₂, so its C-coordinate column is (−1,0)ᵀ.

10. E(x,y)=(−2x+y,x,y) has image W={X+2Y−Z=0}. Is it onto when the codomain is R³? When the codomain is W?

    Show the answer and reasoning

    It is not onto R³; it is onto W.

    Surjectivity asks whether the image equals the declared codomain. W is a proper two-dimensional subspace of R³.

11. Can a linear map from a three-dimensional space to a two-dimensional space be injective?

    Show the answer and reasoning

    No.

    The image dimension is at most 2. Rank-nullity gives kernel dimension at least 3−2=1, so the kernel contains a nonzero vector.

12. For the derivative rule as a map P₂→P₂, give a polynomial outside the image and describe the kernel.

    Show the answer and reasoning

    t² is outside the image; the kernel consists of all constant polynomials.

    The derivative of a polynomial of degree at most 2 has no t² term. Its derivative is zero exactly when its t and t² coefficients vanish.

13. For the shear T and swap R above, which output is (R∘T)(1,2): (2,5) or (4,1)?

    Show the answer and reasoning

    (2,5).

    The rightmost map is applied first: T(1,2)=(5,2), then R(5,2)=(2,5). The other output belongs to T∘R.

What to carry forward

• A basis is an independent spanning list of vectors in a stated space; its order determines the order of coordinates.
• Every finite basis of a space has the same length, its dimension. Ambient coordinate count can be larger.
• A coordinate column names a vector in a chosen basis; reconstruct it with the corresponding basis columns.
• A linear map preserves sums and real scalar multiples. Sending zero to zero alone does not establish linearity.
• Arbitrary images of a basis determine exactly one linear map; images of a redundant list must preserve its relations.
• Column j of [T]ᶜ←ᴮ is [T(bⱼ)]ᶜ. The matrix accepts B coordinates and returns C coordinates.
• The kernel lies in the domain, the image lies in the codomain, and onto means equality with that particular codomain.
• Rank-nullity counts independent lost and retained directions: dim V=dim ker T+dim im T.
• Coordinate changes rename the same vector; compositions apply maps in an order with matching intermediate coordinate conventions.